26-2-2011
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Example
A MgSO4 (FW= 120.4 g/mol) aqueous solution
has a weight fraction of 0.2. What is the
molality of the solution?
Solution
A 0.2 weight fraction means that the solution
contains 20 g MgSO4 and 80 g water (0.080
kg).
Mol MgSO4 = 20 g/(120.4 g/mol) = 0.166 mol
Molality = 0.166 mol/0.080 kg = 2.08 m
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Example
A CaCl2 (FW = 111.0 g/mol) solution is 4.57 m. Find the
mole fraction of CaCl2 and water in solution.
Solution
4.57 m means 4.57 mole CaCl2 in 1000 g of water
Number of moles of water = 1000 g/(18.0 g/mol) = 55.6
mol
XCaCl2 = 4.57 mol/(4.57 + 55.6) = 0.076
Xwater = 55.6/(4.57 + 55.6) = 0.924
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Conversion between Units
1000 g = 1 kg
mass solvent
m = mole/kg
+ add
mass solute
molar
mass
moles solute
mass solution
density
volume solution
1000 mL = 1 L
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M = mole/L
Conversion between conc. units
Calculate the molarity of a 0.396 m glucose
(FW = 180 g/mol) if the density of the
solution is 1.16 g/mL.
Solution
1. First calculate the mass of the solution
2. Use density to change the mass into
volume of solution
3. Calculate M
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0.396 m means 0.396 moles in
1000 g water
• Mass of solution = mass of glucose +
mass of water
• Mass of glucose = mol * FW = 0.396 *180
= 71g
• Mass of solution = 71 + 1000 = 1071g
• Molarity = (0.396 mol glucose/1071 g
soln)* (1.16 g soln/mL soln)* (1000 mL
soln/1 L soln) = 0.429 M
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From M to m
The density of a 2.45 M aqueous methanol (FW =
32g/mol) soln is 0.976 g/mL. What is the
molality of soln?
Solution
1. Using density get the mass of the 1 L solution
2. Get the mass of the solvent by subtracting the
mass of methanol from the mass of the soln.
3. Find the molality
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g methanol = M * FW = 2.45*32 = 78.4g
Mass of 1 L soln = 1000 ml soln * (0.976g
soln/mL soln) = 976 g
Mass of water = 976 – 78.4 = 898 g
m = 2.45 mol methanol/0.898 kg water) =
2.73 m
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What is the molality of a 5.86 M ethanol (C2H5OH)
solution whose density is 0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
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=
5.86 moles C2H5OH
= 8.92 m
0.657 kg solvent
12.3
Convert % mass to Molarity
What is the Molarity of a 95% acetic acid
solution? (density = 1.049 g/mL)
If you assume 1 L, that amount of solution = 1049 g
95% of the solution is acetic acid
1049 g solution x 0.95 = 997 g solute
997 g X 1 mol/60.05 g = 16.6 mol solute
Since we assumed 1 L, that’s 16.6 mol / 1 L or
16.6 M
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Effect of pressure on solubility
Liquids and solids exhibit practically no change of
solubility with changes in pressure. Gases as
might be expected, increase in solubility with an
increase in pressure. Henry's Law states that:
The solubility of a gas in a liquid is directly
proportional to the pressure of that gas above
the surface of the solution.
If the pressure is increased, the gas molecules are
"forced" into the solution since this will best
relieve the pressure that has been applied.
The number of gas molecules above solution is
decreased.
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The solubility of a gas (Cg in mol/L) in a liquid is
directly proportional to the pressure of the gas
(in atm) above the solution.
Cg = kg Pg
kg is the Henry's law constant. This relation is also
useful in calculation of the solubility of a gas at
one pressure provided that the solubility at some
other pressure is known:
Cg1 / Cg2 = Pg1 / Pg2
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Example
The solubility of nitrogen gas at 25 oC and 1
atm is 6.8*10-4 M. What is the
concentration of nitrogen dissolved in
water under atmospheric conditions? The
partial pressure of nitrogen in the
atmosphere is 0.78 atm.
Solution
First find Henry’s constant:
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Cg = kg Pg
6.8*10-4 = kg * 1
6.8*10-4 = kg
The solubility of nitrogen gas can then be
calculated:
Cg = kg Pg
Cg = 6.8*10-4 * 0.78
Cg = 5.3*10-4 M
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Example
At 25 oC, O2 gas collected over water at a
total pressure of 1.0 atm is soluble to the
extent of 0.0393 g/L. What would its
solubility be if its partial pressure over
water was 800 torr. Pwater = 23.8 torr at 25
oC
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PT = PO2 + Pwater
760 torr = PO2 + 23.8
PO2 = 736 torr
Substitution in the equation Cg1 / Cg2 = Pg1 / Pg2
gives:
0.0393/CO2 = 736/800
CO2 = 0.0427 g/L
For Henry’s law to be valid, the gas should not
react with solvent.
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